# Sum of radicals greater than 1

Prove that for every $n,m \in \Bbb N$ $$\frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1$$

According to Bernoulli's inequality $$(1+m)^\frac{1}{n} \leq 1+\frac{m}{n}, \\ (1+n)^\frac{1}{m} \leq 1+\frac{n}{m},$$ so that $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geq \frac{1}{1+\frac{m}{n}}+\frac{1}{1+\frac{n}{m}}=1.$$
For every $a,b \in \mathbb{N}$ we have $$\left(1+\frac{b}{a}\right)^a=\sum_{i=0}^a{a\choose i}\left(\frac{b}{a}\right)^i=1+b+\ldots \ge 1+b.$$ Therefore $$(1+b)^{1/a}\le 1+\frac{b}{a}=\frac{a+b}{a} \quad \forall a,b \in \mathbb{N}.$$ Hence for every $m,n \in \mathbb{N}$ we have $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \ge \frac{n}{m+n}+\frac{m}{m+n}=1.$$